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<h3 class="heading"><span class="type">Paragraph</span></h3>
<p>Consider the general case</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_10.html">
\begin{equation}
y^{\prime}+p(x) y=q(x).\tag{2.1.6}
\end{equation}
</div>
<p class="continuation">We multiply it by the integrating factor <span class="process-math">\(u(x)\)</span> (yet to be determined) and have</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_10.html">
\begin{equation}
u(x) y^{\prime}+u(x) p(x) y=u(x) q(x).\tag{2.1.7}
\end{equation}
</div>
<p class="continuation">We want <span class="process-math">\(u(x)\)</span> to be such a function that</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_10.html">
\begin{equation}
\begin{aligned}
&amp;u(x) y^{\prime}+u(x) p(x) y=\frac{\textrm{d}}{\textrm{d} x} (u(x) y)=u(x) y^{\prime}+\frac{\textrm{d} u(x)}{\textrm{d} x} y,\\
&amp;\rightarrow \frac{1}{u}\frac{\textrm{d} u}{\textrm{d} x}=p(x) \rightarrow \frac{\textrm{d}}{\textrm{d} x} \ln |u(x)|=p(x) \rightarrow \ln |u(x)|=\int p(x) \textrm{d} x+k,
\end{aligned}\tag{2.1.8}
\end{equation}
</div>
<p class="continuation">where <span class="process-math">\(k\)</span> is a constant. The simplest choice is to set <span class="process-math">\(k=0\text{.}\)</span> Then</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_10.html">
\begin{equation}
|u(x)|=e^{\int p(x) \textrm{d} x}.\tag{2.1.9}
\end{equation}
</div>
<p class="continuation">Taking <span class="process-math">\(u(x)\)</span> to be non-negative, we have</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_10.html">
\begin{equation}
u(x)=e^{\int p(x) \textrm{d} x}.\tag{2.1.10}
\end{equation}
</div>
<p class="continuation">Equation <a href="" class="xref" data-knowl="./knowl/eq2_10.html" title="Equation 2.1.10">(2.1.10)</a> gives one form of integrating factor.</p>
<span class="incontext"><a href="sec2_1.html#p-18" class="internal">in-context</a></span>
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